Timeline for $f: [0,1]\rightarrow L^1(\Omega)$ as a (measurable?) function from $[0,1]\times \Omega\rightarrow \mathbb{R}$
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 9, 2023 at 21:27 | comment | added | Akira | @KeeferRowan I think the intended interpretation is that there is at least one such measurable representation. Dear professor Edgar, could you confirm if my understanding is correct? | |
| Jun 21, 2021 at 19:37 | comment | added | Keefer Rowan | I think the claim that (2) is clear can't be quite right. Let $f : [0,1] \to L^1(\Omega)$ and call $r : [0,1] \times \Omega \to \mathbb{R}$ a representation of $f$ if for (almost) every $t \in [0,1]$, $f(t) = r(t,\cdot)$ a.e. (this is as uniquely as we can define things as $L^1(\Omega)$ is only defined up to a.e. equivalence). Then one can show there are non-measurable representations of $f$ even if $f(t) =0$ (say with $\Omega = (0,1)$). Your comment suggests to me that this shouldn't be possible, but I may be misinterpreting. | |
| Mar 9, 2018 at 18:12 | history | edited | LSpice | CC BY-SA 3.0 |
Added links to papers and MSN
|
| Mar 9, 2018 at 15:16 | vote | accept | Xiao | ||
| Mar 9, 2018 at 11:46 | history | answered | Gerald Edgar | CC BY-SA 3.0 |